I am trying to make photo gallery.
I got stuck as many times, again.
here i am.
trying to output categoty name:
include ('config.inc.php');
include "portopager.php";
$sql="Select * from gallery_photos photo_category where photo_category='".$_GET[kategorija]."'";
$pclass = new Pager;
$limit = 3; //per page
$start = $pclass->findStart($limit);
$count = mysql_num_rows(mysql_query($sql));
$pages = $pclass->findPages($count, $limit);
$sql.=" LIMIT ".$start.", ".$limit;
$queryp=mysql_query($sql);
?>
<center>
<table>
<tr>
<td align=center colspan="<?PHP echo $per_row; ?>"><?PHP echo $table_header; ?></td></tr><tr>
<tr>
<td align=center colspan="<?PHP echo $per_row; ?>">
<?
$sql = "SELECT * FROM gallery_category ORDER BY category_id";
$rezultati = mysql_query($sql) or die(mysql_error());
while ($polje = mysql_fetch_array($rezultati)) {
$kategorija1 = $polje['category_name'];
$kategorija = $polje['category_name'];
echo '| <a href="gal.php?kategorija='.$polje[ 'category_id' ].'">'.$polje['category_name'].'</a> | ';
}
?>
</td>
</tr>
<?PHP
if (mysql_num_rows($queryp) == 0)
echo "0 Photos";
elseif (mysql_num_rows($queryp) == 1){
}
$kategorija = $polje['category_name'];
$ixi = 0;
while ($rowp=mysql_fetch_array($queryp)){
if ($ixi < $limit) {
if (($ixi % $per_row) == 0) {
echo "</tr><tr>";
}
echo "<td>";
$image_name = "./photos/"."/tb_".$rowp['photo_filename'];
$image_name2 = "./photos/" . $rowp['photo_filename'];
$comments = $rowp['photo_caption'];
cell();
echo "</td>";
$ixi ++;
}
}
echo "</tr></table></center>";
echo "<center>";
echo "</table><br>\n";
echo $pagelist = $pclass->pageList($_GET['page'], $pages, $_GET['kategorija']);
?>
you access gal.php, it displays all categories. like, 2000, 2001, 2002
now when you click on one of them, like gal.php?kategorija=1
2000=1
2001=2
2002=3
as you can see from my code i am using pager class, what i would like to output category name , on either right site of pagination or left.
<<123> 2000
Can any1 help ?
Thank you
